Which integral gives the length of the graph of $f(t)=\dfrac{1-2t}{t^2}$ between $t=a$ and $t=b$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_a^b\sqrt{1+\bigg(\dfrac{-2-2t}{t^3}\bigg)^2}~dt$ (Choice B) B $ \int_a^b\sqrt{1+\dfrac{2-2t}{t^3}}~dt$ (Choice C) C $ \int_a^b\sqrt{1+\bigg(\dfrac{2t-2}{t^3}\bigg)^2}~dt$ (Choice D) D $ \int_a^b\sqrt{1+\dfrac{2t-2}{t^3}~}~dt$
Answer: Recall that the formula for arc length of $~f(t)~$ over $~[a, b]~$ is $ L=\int_a^b\sqrt{1+\big[f\,^\prime(t)\big]^2}~dx\,$. First calculate $~f\,^\prime(t)$ and simplify. $ f\,^\prime(t)=\dfrac{t^2(-2)-(1-2t)\cdot2t}{(t^2)^2}=\dfrac{2t^2-2t}{t^4}=\dfrac{2t-2}{t^3}$ Next, use the formula above to write the integral expression that gives the arc length in question. $L=\int_{a}^b\sqrt{1+\bigg(\dfrac{2t-2}{t^3}\bigg)^2}~dt$